# Vertical motion initial velocity given max height

I'm having a hard time finding a formula that calculates the initial vertical velocity for a projectile that should reach a certain height.

I'm using this formula to calculate both vertical and horizontal displacement:

This works great if I have the initial velocity, with any angle (even for motion in vertical axis only!).

Problem is that I have maximum height and angle, and I need to calculate that initial velocity. I tried the following formula:

And it works for angles other than 90 degrees (PI / 2). But I need to handle the case of vertical motion only. This angle works in the previous formula, but not here. Also obvously, I need to handle the other angles too (for which case this formula works good).

Is there any other formula I can try?

I have tried Wikipedia and Khan Academy, but couldn't find what I needed.

• Use $v^2 = u^2 + 2as$. The velocity $v$ is $v_0\sin\theta$. – John Rennie Oct 30 '14 at 7:08
• @JohnRennie, what are $u$, $a$ and $s$? – Veehmot Oct 30 '14 at 7:13
• @Veehmot $u$ is initial velocity ($v_0 sin \theta$), $a$ is acceleration, and $s$ is distance. If you like: $$(v_f \space sin\theta)^2 = (v_i \space sin\theta)^2 + 2ax$$ – Goodies Oct 30 '14 at 7:36
• At the top of the trajectory the vertical component of velocity is zero, so $v = 0$. Your equation simplifies to $0 = (v_0\sin\theta)^2 + 2gh$. Note that $g$ is negative. – John Rennie Oct 30 '14 at 7:58
• And since for this particular case (vertical movement), the angle is 90, thus $\sin\theta = 1$, the equation should look like $0 = {v_{0}}^{2} + 2gh$, is this correct? – Veehmot Oct 30 '14 at 8:07

$$v_{0} = \sqrt{2gh}$$